Question

Help please find the derrative

Answer 1

cot inverse ( t) + cot inverse ( 1/t)

Answer 2

anyone can help me

Answer 3

hold the mayo

Answer 4

@Directrix

Answer 5

im so lost about this ..

Answer 6

hmmm one sec... not one I had offhand =)
but \(\cfrac{d}{dx}[cot^{-1}(x)]=-\cfrac{1}{1+x^2}\qquad thus
\\ \quad \\
cot^{-1}(t)+cot^{-1}\left( \frac{1}{t} \right)\implies cot^{-1}(t)+cot^{-1}\left( t^{-1} \right)
\\ \quad \\
-\cfrac{1}{1+t^2}-\cfrac{1}{1+t^{-1}}\cdot -1t^{-2}
\\ \quad \\
-\cfrac{1}{1+t^2}-\cfrac{1}{1+t^{-1}}\left( -\cfrac{1}{t^2}\right)\)

Answer 7

on the second term, you'd do the chain-rule, is all

Answer 8

hmm can you use qoutient rule

Answer 9

well, you don't have a divisor or denominator, so hmm well.. on the second term? yes, you can, you'd end up in the same spot anyway

Answer 10

oh hmmm okay so then

Answer 11

(1/t^2) ^2 isnt that that 1/t^2

Answer 12

hmm actually rats... yea, missed that part hmmm lemme redo it some

Answer 13

ok lol

Answer 14

brb..

Answer 15

k

Answer 16

\(\cfrac{d}{dx}[cot^{-1}(x)]=-\cfrac{1}{1+x^2}\qquad thus
\\ \quad \\
cot^{-1}(t)+cot^{-1}\left( \frac{1}{t} \right)\implies cot^{-1}(t)+cot^{-1}\left( t^{-1} \right)
\\ \quad \\
-\cfrac{1}{1+t^2}-\cfrac{1}{1+(t^{-1})^2}\cdot -1t^{-2}\implies
-\cfrac{1}{1+t^2}-\cfrac{1}{1+\frac{1^2}{t^2}}\left( -\cfrac{1}{t^2} \right)
\\ \quad \\
-\cfrac{1}{1+t^2}-\cfrac{1}{\frac{t^2+1}{t^2}}\left( -\cfrac{1}{t^2} \right)\implies
-\cfrac{1}{1+t^2}-\cfrac{t^2}{t^2+1}\left( -\cfrac{1}{t^2} \right)
\\ \quad \\
-\cfrac{1}{1+t^2}+\cfrac{\cancel{t^2}}{(t^2+1)\cancel{t^2}}\implies
-\cfrac{1}{t^2+1}+\cfrac{1}{t^2+1}\implies \cfrac{\cancel{-1+1}}{t^2+1}\)